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3 years ago

6

A man pulls a sled at a constant velocity across a horizontal snow surface. if a force of 80 n is being applied to the sled rope

at an angle of 53Â° to the ground, what is the force of friction between sled and snow?

2 answers:

algol [13]3 years ago

4 0

Answer:

The force of friction between sled and snow is 48.14 N

Explanation:

It is given that,

Force applied on the sled, F = 80 N

The angle with the sled and the ground is 53 degrees. We need to find the force of friction between sled and snow.

The horizontal force acting in the horizontal direction is given by :

$F_x=F\ cos\theta$

$F_x=80\times cos(53)$

$F_x=48.14\ N$

So, the force of friction between sled and snow is 48.14 N. Hence, this is the required solution.

balandron [24]3 years ago

3 0

A constant velocity implies the two forces must be equal and opposite.
Friction acts horizontal to the ground, therefore we must find the force applied to the sled rope that acts horizontal to the ground.
Do this by resolving:
Force = 80cos53
The force opposing this is equal, and so also = 80cos53 = 48 N (2 sig. fig.)

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Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0

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Answer:

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Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

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However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

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$v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}$

you replace this value of v in the equation (1). Also, you replace the values of r and m:

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Vanyuwa [196]

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An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro

Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

$\sum F = m_b a$

$m_Bg-T_B = m_Ba$

$T_B = m_Bg-m_Ba$

In the case of mass A,

$\sum F = m_A a$

$T_A = m_Ag-m_Aa$

Making summation of Torques in the Pulley we have to

$\sum\tau = I\alpha$

$T_BR_0-T_AR_0=I\alpha$

$T_B-T_A=I\frac{a}{R^2_0}$

Replacing the values previously found,

$(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}$

$(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}$

$a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}$

$a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}$

$a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}$

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PART B) Ignoring the moment of inertia the acceleration would be given by

$a' =\frac{(m_B-m_A)g}{(m_B+m_A)}$

$a' =\frac{(8-6.8)(9.8)}{(8+6.8)}$

$a' = 0.7945$

Therefore the error would be,

$\%error = \frac{a'-a}{a}*100$

$\%error = \frac{0.7945-0.7736}{0.7736}*100$

$\%error = 2.7%$

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3 years ago